2017-11-24T23:38:11ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=22747Whether 0⁰ is defined or not, it is convenient in many mathematical formulas to treat it as equal to 1. For instance, consider the cosine power series:

If you put x = 0, you’ll find that the first term involves the expression 0⁰. Rather than saying the formula is invalid because 0⁰ is undefined, we happily let 0⁰ = 1 anyway.

]]>http://www.mathisfunforum.com/profile.php?id=2060892017-08-30T16:53:27Zhttp://www.mathisfunforum.com/viewtopic.php?pid=400609#p400609It's more appropriate to say 0^0 is 1 than saying its undefined ]]>http://www.mathisfunforum.com/profile.php?id=2165852017-08-30T16:36:25Zhttp://www.mathisfunforum.com/viewtopic.php?pid=400608#p400608Yeah. It's better to use 0^0 as 1 rather than undefined. ]]>http://www.mathisfunforum.com/profile.php?id=1957442016-02-12T14:01:50Zhttp://www.mathisfunforum.com/viewtopic.php?pid=375517#p375517hi NakulG

Numbers have the value that is defined for them. For counting numbers it is straight forward, but as you delve into the number system it becomes harder to decide on a value.

Most mathematicians work with the definition zero to the power zero equals 1. This maintains consistency across several theorems and gives completeness.

Bob

]]>http://www.mathisfunforum.com/profile.php?id=676942016-01-01T15:28:06Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373275#p373275It's actually quite enlightening to graph the various functions and see why it's so ambiguous what happens at 0.

For 0^0, if you graph:x^x you get 1, x^0 you get 1, and 0^x you get 0.

For 0/0, if you graph: x/x you get 1, and 0/x you get 0.

For ordinary division by 0, if you graph y/x you get plus or minus infinity. Plus the implication that if, for instance 1/0 = 2/0, then 1 = 2

]]>http://www.mathisfunforum.com/profile.php?id=2120902016-01-01T12:39:56Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373263#p373263Thanks for your response.I think, equating 0^0 to anything is our construct, I think we should leave 0^0 as it is without equating it to anything that we understand today.]]>http://www.mathisfunforum.com/profile.php?id=2080012016-01-01T07:38:11Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373257#p373257Hi NakulG, Somebody else is free to correct me, but I don't think there is any practical use for an answer to 0^0, even in quantum theory ... except for making certain theorems more elegant, as mentioned, if it equals 1.

I don't think your second point is gibberish at all. One way of thinking about exponentiation is 'multiply the base by itself exponent number of times", and an extension to the zeroth power is often given as "It is not multiplied by itself at all, so we are just left with the empty product", which is 1. I.e. 0^2 = 1*0*0, 0^1 = 1*0, 0^0 = 1.The problem is that you can't prove this using the laws of exponentiation, or by using limits, so it's more of an answer for the sake of convenience. The problem is that that answer requires dividing by the base, and division by zero is not allowed (how can we say that 1*0 / 0 = 1?)

As 0^0 = 0/0, I suppose answering what it means to ask about one implies what it means to ask about the other.

]]>http://www.mathisfunforum.com/profile.php?id=2120902016-01-01T01:02:13Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373233#p373233For one thing we would have to modify the binomial theorem and some other series. It is more convenient to define it like this 0^0 = 1.]]>http://www.mathisfunforum.com/profile.php?id=337902015-12-31T15:12:13Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373189#p373189Thanks for all the responses, these are very insightful.But1. why would a 0^0 situation arise, that it would need to be answered.2. 0 represents the 'lack of anything'. So 'lack of anything' times 'lack of anything' would be something or even a set or is this gibberish. That is why I wanted to know what is it to try to answer 0^0?]]>http://www.mathisfunforum.com/profile.php?id=2080012015-12-31T14:42:19Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373184#p373184But... 0 ^ 0 = 0 ^ (1 - 1) = (0 ^ 1) / (0 ^ 1) = 0/0]]>http://www.mathisfunforum.com/profile.php?id=2120312015-12-31T13:14:20Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373183#p3731830^0 is the empty product, which is 1 by definition]]>http://www.mathisfunforum.com/profile.php?id=959042015-12-31T09:35:37Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373177#p373177The link bobbym posted covers this quite well. For simplicity we take 0^0 = 1, although it is really undefined.

Relentless wrote:

This is going to sound really wacky and unschooled, but if there are two limits that predict 0^x = 0 if x = 0 and y^0 = 1 if y = 0, instead of throwing one's hands up, could they average the two?

In the context of limits of functions, I've only seen this practice adopted when dealing with jump discontinuities in Fourier series, in which one takes the average of the limits from the left and right at a jump discontinuity. However, a similar idea has been used with sequences, called Cesaro summation.

]]>http://www.mathisfunforum.com/profile.php?id=2060892015-12-31T09:07:07Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373174#p373174This is going to sound really wacky and unschooled, but if there are two limits that predict 0^x = 0 if x = 0 and y^0 = 1 if y = 0, instead of throwing one's hands up, could they average the two?]]>http://www.mathisfunforum.com/profile.php?id=2120902015-12-31T05:03:39Zhttp://www.mathisfunforum.com/viewtopic.php?pid=373153#p373153